Question 77060
a)

First factor out the negative 1

{{{y=-1(x^2-6x+8)}}}

*[invoke completing_the_square 1, -6, 8]

So if we multiply everything by -1 we get

{{{y=-(x-3)^2+1}}}

So the vertex is (3,1)


b)
The line of symmetry will go through the vertex so the equation is
x=3


c)
The range is any value y can be, so by our graph we can see the range is
(*[Tex \large -\infty], 1]

Here's our graph

{{{ graph( 300, 200, -6, 5, -10, 10, -x^2+6x-8) }}} graph of {{{y=-(x-3)^2+1}}}